Question

Solve the inequation $\frac{2x-3}{4}+9\ge 3+\frac{4x}{3}$. Also represent the solution on number line.

Answer 1

We have,$\frac{2x-3}{4}+9\ge 3+\frac{4x}{3}$.

$\Rightarrow \frac{2x-3}{4}-\frac{4x}{3}\ge 3-9$.

$\Rightarrow \frac{6x-9-16x}{12}\ge -6$

$\Rightarrow -10x-9\ge -72$

$\Rightarrow -10x\ge -72+9$

$\Rightarrow -10x\ge -63\Rightarrow x\le \frac{63}{10}$

$\Rightarrow x\in (-\infty ,\frac{63}{10})$

Hence, the solution set of the given inequation is $x\in (-\infty ,\frac{63}{10})$.

The representation of solution set on number line is shown below.